Question: Multiply the following complex numbers: $({3-4i}) \cdot ({1+i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3-4i}) \cdot ({1+i}) = $ $ ({3} \cdot {1}) + ({3} \cdot {1}i) + ({-4}i \cdot {1}) + ({-4}i \cdot {1}i) $ Then simplify the terms: $ (3) + (3i) + (-4i) + (-4 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 3 + (3 - 4)i - 4i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 3 + (3 - 4)i - (-4) $ The result is simplified: $ (3 + 4) + (-1i) = 7-i $